Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $\displaystyle x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.
Thanks!
(a) is a direct application of the cosine rule: $\displaystyle 10^2 = 7^2 + x^2 - 2 (x)(7) \cos 50^0$.
(b) $\displaystyle \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)
$\displaystyle \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)
$\displaystyle \frac{7}{\sin \beta} = \frac{10}{\sin 50^0}$ .... (2)
From (2) get $\displaystyle \beta$. From $\displaystyle \alpha + \beta + 50^0 = 180^0$ get $\displaystyle \alpha$. Now use (1) to solve for $\displaystyle x$.
Huh? Just use the quadratic formula:
$\displaystyle \begin{aligned}
x^2 - 14(\cos 50^{\circ}) x - 51 &= 0 \\
x^2 - 9x - 51 &\approx 0 \\
x &\approx \frac{9 \pm \sqrt{(-9)^2 - 4(1)(-51)}}{2(1)} \\
&\approx \frac{9 \pm \sqrt{284.98}}{2} \\
\end{aligned}$
$\displaystyle x \approx 12.94$
$\displaystyle x \approx -3.94$
Reject the 2nd solution, so x is about 13 cm.
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